Properties of the Cosine Function

Example 1: Simplify the expression \(\sin^4 x + \sin^2 x + 2\sin^2 x\cos^2 x + 2\cos^2 x + \cos^4 x\).

Solution: There are subtle ways to simplify this. Notice that the first three terms have a common factor of \(\sin^2 x\), and the last three terms have a common factor of \(\cos^4 x\). We can break up the middle term so we can begin factoring by grouping:

$$\sin^4 x + \sin^2 x + \sin^2 x\cos^2 x + \sin^2 x\cos^2 x + 2\cos^2 x + \cos^4 x$$

Factor a \(\sin^2 x\) out of the first three terms and a \(\cos^2 x\) out of the last three terms:

$$\sin^2 x(\sin^2 x + 1 + \cos^2 x) + \cos^2 x(\sin^2 x + 2 + \cos^2 x)$$

Rearrange the terms inside the parentheses a little bit to see something interesting:

$$\sin^2 x(\sin^2 x + \cos^2 x + 1) + \cos^2 x(\sin^2 x + \cos^2 x + 2)$$

The expression \(\sin^2 x + \cos^2 x\) pops up twice in this larger expression, which, according to Pythagoras's Identity, is equal to \(1\). Substitute that into the expression and simplify:

$$\sin^2 x(1 + 1) + \cos^2 x(1 + 2) = 2\sin^2 x + 3\cos^2 x$$

That really simplified the expression, but we can keep going. The \(3\cos^2 x\) can be broken up in a similar way to simplify this further. In general, squares of sines and cosines are a sign that Pythagoras's Identity can be used:

$$2\sin^2 x + 2\cos^2 x + \cos^2 x$$

Factor a \(2\) out of the first two terms:

$$2(\sin^2 x + \cos^2 x) + \cos^2 x$$

Pythagoras's Identity can be used again:

$$2(1) + \cos^2 x = 2 + \cos^2 x$$

Example 2: Solve for \(\cos x\) in Pythagoras's Identity, \(\sin^2 x + \cos^2 x = 1\).

Solution: Isolate the cosine term by subtracting \(\sin^2 x\) from both sides of the equation:

$$\cos^2 x = 1 - \sin^2 x$$

Take the square root of both sides of the equation:

$$\cos x = \pm \sqrt{1 - \sin^2 x}$$

Example 3: Show that the equation \(\sqrt{8 - \sin^2 x} = 4 + \cos x\) has no real solutions.

Solution: The square root on the left side of the equation indicates that substituting \(\cos x\) out for \(\sqrt{1 - \sin^2 x}\):

$$\sqrt{8 - \sin^2 x} = 4 + \sqrt{1 - \sin^2 x}$$

Trying to solve this equation by squaring won't be as messy if we let \(\sin^2 x = a\), because the exponents in the equation will be lower.

$$\sqrt{8 - a} = 4 + \sqrt{1 - a}$$

Now, square the equation:

$$8 - a = 16 + 8\sqrt{1 - a} + 1 - a$$

This can be simplified:

$$8 - a = 17 + 8\sqrt{1 - a} - a \Rightarrow$$ $$-9 - a = 8\sqrt{1 - a} - a \Rightarrow$$ $$-9 = 8\sqrt{1 - a}$$

Now isolate \(a\) by dividing both sides of the equation by dividing both sides of the equation by \(8\):

$$\frac{-9}{8} = \sqrt{1 - a}$$

Now square both sides of the equation again:

$$\frac{81}{64} = 1 - a$$

Now solve for \(a\):

$$\frac{17}{64} = -a \Rightarrow$$ $$a = -\frac{17}{64}$$

This means that \(\sin^2 x = -\frac{17}{64}\). If we were to solve for \(x\), we would need to take the square root of both sides of the equation, but that would give a complex number on the right hand side, so \(x\) cannot have any real solutions, because the arcsine of a complex number does not create a real number.

Example 4: Prove the identity \(\cos(2a) =\cos^4(a) - \sin^4(a)\).

Solution: Notice that the left hand side of this equation is the same as the left hand side in the double angle identity, so we should try to prove that

$$\cos^4(a) - \sin^4(a) = \cos^2(a) - \sin^2(a)$$

This is actually pretty easy, because the left hand side is a difference of squares, so it can be factored as

$$\cos^4(a) - \sin^4(a) = (\cos^2(a) + \sin^2(a))(\cos^2(a) - \sin^2(a))$$

The first factor, according to Pythagoras's Identity, is just equal to \(1\), so

$$\cos^4(a) - \sin^4(a) = \cos^2(a) - \sin^2(a)$$

So as a result, since \(\cos(2a) = \cos^2(a) - \sin^2(a)\), it is also true that \(\cos(2a) = \cos^4(a) - \sin^4(a)\).

Example 5: Prove that \(1 - 2\sin^2(2a) = \cos(3a)\sin(a) - \cos(a)\sin(3a)\).

Solution: Notice that by the cosine addition formula:

$$\cos(3a)\sin(a) - \cos(a)\sin(3a) = \cos(3a + a) = \cos(4a)$$

Also, \(\cos(4a) = \cos(2 \cdot 2a)\). Therefore, the double-angle identity can be used on this expression to give

$$\cos(4a) = \cos(2 \cdot 2a) = 1 - 2\sin^2(2a)$$

Therefore \(1 - 2\sin^2(2a) = \cos(3a)\sin(a) - \cos(a)\sin(3a)\).

Example 6: If \(0 \leq y \leq 2\pi\), solve the equation \(2\cos(2y) - \sin^2(2y) = 2\).

Solution: These types of equations are easier to solve if they are only in terms of one trigonometric function. The term \(-\sin^2(2y)\) suggests we should use cosines because

$$1 - \sin^2(2y) = \cos^2(2y)$$

To create that second cosine, add \(1\) to both sides of the equation:

$$2\cos(2y) + 1 - \sin^2(2y) = 3 \Rightarrow$$ $$\cos^2(2y) + 2\cos(2y) = 3$$

Factoring is a possibility here, in particular completing the square, because that will give us only one cosine term. Since

$$\cos^2(2y) + 2\cos(2y) + 1 = (\cos(2y) + 1)^2$$

We can complete the square by adding \(1\) to both sides of the equation:

$$\cos^2(2y) + 2\cos(2y) + 1 = 4$$

Now write the left side with the square:

$$(\cos(2y) + 1)^2 = 4$$

Take the square root of both sides of the equation:

$$\cos(2y) + 1 = \pm 2$$

Before we proceed, notice that the range of \(\cos(2y)\) is \([-1, 1]\), so if \(\cos(2y) + 1 = -2\), then \(\cos(2y) = -3\), and we cannot have any solutions for \(y\) in this case because \(\arccos(3)\) is undefined. Therefore:

$$\cos(2y) + 1 = 2 \Rightarrow$$ $$\cos(2y) = 1$$

Now use the arccosine function to find the possible values of \(y\):

We list the values of \(2y\) in the range \([0, 4\pi]\) because dividing this range by \(2\) gives \(y \in [0, 2\pi]\). $$2y = 0, 2\pi, 4\pi \Rightarrow$$ $$y = 0, \pi, 2\pi$$

Example 7: Prove that

$$\cos(11\theta) = \sin(17\theta) - 2\sin^2(14\theta)\sin(17\theta) + 2\sin(14\theta)\cos(14\theta)\cos(17\theta)$$

Solution: While this equation looks messy, it can be simplified pretty quickly. First recall the double-angle identity for sine, which states that

$$\sin(2\theta) = 2\sin(\theta)\cos(\theta)$$

As a result, we can simplify the last term on the right side of the equation:

$$2\sin(14\theta)\cos(14\theta)\cos(17\theta) = \sin(2 \cdot 14\theta)\cos(17\theta) = \sin(28\theta)\cos(17\theta)$$

Therefore the equation becomes

$$\cos(11\theta) = \sin(17\theta) - 2\sin^2(14\theta)\sin(17\theta) + \sin(28\theta)\cos(17\theta)$$

Now, we can extract another identity from this expression by noticing that the first two terms have a common factor of \(\sin(17\theta)\). Factoring that out of those two terms gives

$$\cos(11\theta) = \sin(17\theta)(1 - 2\sin^2(14\theta)) + \sin(28\theta)\cos(17\theta)$$

The second factor is one equivalent form of \(\cos(28\theta)\) by the double-angle identity for cosine. Since

$$\cos(2\theta) = 1 - 2\sin^2(\theta)$$

we can substitute \(\theta \rightarrow 14\theta\) and claim that

$$\cos(28\theta) = 1 - 2\sin^2(14\theta)$$

Therefore,

$$\cos(11\theta) = \sin(17\theta)\cos(28\theta) + \sin(28\theta)\cos(17\theta)$$

Actually, this is the angle subtraction formula in disguise, since

$$\cos(11\theta) = \cos(28\theta - 17\theta)$$

Therefore this identity is true by the angle subtraction formula.

Example 8: If \(0 \leq x \leq \pi\), solve the equation

$$2(\cos(x) + 1)(\cos(x) - 1) = -\frac{1}{2}$$

Solution: The two factors with cosine in them represent a difference of squares. Expand:

$$2(\cos^2(x) - 1) = -\frac{1}{2}$$

Distribute the \(2\):

$$2\cos^2(x) - 2 = -\frac{1}{2}$$

Now, we can get rid of the square since \(2\cos^2(x) - 1 = \cos(2x)\). To transform the left side accordingly, add \(1\) to both sides of the equation:

$$2\cos^2(x) - 1 = \frac{1}{2}$$

Now use the aforementioned identity:

$$\cos(2x) = \frac{1}{2}$$

Now solving for cosine is straightforward. Take the arccosine of both sides of the equation:

$$2x = \frac{\pi}{3}, \frac{5\pi}{3}$$

Since \(x\) is in the range \([0, \pi]\), we double the length of the range for the possible values of \(2x\), those within \([0, 2\pi]\). Therefore

$$x = \frac{\pi}{6}, \frac{5\pi}{6}$$

Example 9: Find the derivative of \(\cos(6x^2 - 7x + 5)\).

Solution: We must use the Chain Rule to differentiate this, where the insider function is \(6x^2 - 7x + 5\) and the outside function is cosine. We get

$$-\sin(6x^2 - 7x + 5) \cdot \frac{d}{dx}(6x^2 - 7x + 5) = -(12x - 7)\sin(6x^2 - 7x + 5)$$

Example 10: Find the derivative of \(\cos(3y)\sin(5y) - \cos(5y)\sin(3y)\).

Solution: We could brute-force the differentiation with the Product Rule and Chain Rule, but instead realize that this can be simplified greatly with the cosine angle addition identity,

$$\cos(a + b) = \cos(a)\sin(b) - \cos(b)\sin(a)$$

So, this expression equals

$$\cos(5y + 3y) = \cos(8y)$$

Now, where the outside function is cosine and the inside function is \(8y\), we differentiate with the Chain Rule:

$$\frac{d}{dy}(\cos(8y)) = -\sin(8y) \cdot \frac{d}{dy}(8y) = -8\sin(8y)$$

Example 11: Find the slope of the tangent line of \(y\cos x + x^2 = 77\) when \(x = 0\).

Solution: We must first find \(\frac{dy}{dx}\). To do so, just differentiate the entire equation remembering that \(y\) is a function of \(x\) and the Chain Rule must be applied on it. \(\frac{d}{dx}(y) = \frac{dy}{dx}\).

$$\frac{dy}{dx}\cos(x) - y\sin(x) + 2x = 0$$

Now we solve for \(\frac{dy}{dx}\) in this equation.

$$\frac{dy}{dx}\cos(x) = y\sin(x) - 2x$$

Divide both sides of the equation by \(\cos(x)\):

$$\frac{dy}{dx} = \frac{y\sin(x) - 2x}{\cos(x)}$$

We can plug \(x = 0\) into this derivative to find the slope of the tangent line at \(x = 0\):

$$\frac{dy}{dx} = \frac{y\sin(0) - 2(0)}{\cos(0)} = \frac{y(0) - 0}{1} = 0$$

Example 12: Prove that the \(4n\)th derivative of \(\cos(x)\) is \(\cos(x)\), where \(n\) is a positive integer.

Solution: If this statement is claimed to be true for all positive integers \(n\), then we can prove that it is true for \(n = 1\).

First derivative:

$$\frac{d}{dx}(\cos(x)) = -\sin(x)$$

Second derivative:

$$\frac{d}{dx}(-\sin(x)) = -\cos(x)$$

Third derivative:

$$\frac{d}{dx}(-\cos(x)) = \sin(x)$$

Fourth derivative:

$$\frac{d}{dx}(\sin(x)) = \cos(x)$$

Thus the statement is true for \(n = 1\). Differentiating this expression four more times gives the same result because the calculations are exactly the same (giving \(n = 2\), then four more differentiations returns the function to \(\cos(x)\), and so on infinitely. Therefore the statement holds true.

Example 13: Integrate \(\int{\cos(4x - 1)dx}\).

Solution: We perform the Chain Rule backwards. The outside function is \(\cos(x)\), so the antiderivative converts this to a sine of the integrand's argument. Since the inside function is \(4x - 1\), we must divide the derivative of this function (whereas in forward differentiation we multiply it). This gives an antiderivative of

$$\frac{1}{4}\sin(4x - 1) + C$$

If you are new to this integration technique, feel free to double-check this procedure by differentiating this function. You should get the original integrand.

Example 14: Integrate \(\int{\frac{\sin(x)}{1 - \sin^2(x)}dx}\).

Solution: Here we apply Pythagoras's Identity. Since \(\sin^2(x) + \cos^2(x) = 1\), we know that

$$1 - \sin^2(x) = \cos^2(x)$$

Therefore the integrand becomes

$$\int{\frac{\sin(x)}{\cos^2(x)}dx}$$

The derivative of cosine is negative sine, so we let \(u = \cos(x)\).

$$\int{\frac{\sin(x)}{u^2}dx}$$

Since \(\frac{d}{dx}(\cos(x)) = -\sin(x)\), we can multiply the integrand and outside the integrand by \(-1\) to get a negative sign into the integrand:

$$-\int{\frac{-\sin(x)}{u^2}dx}$$

How does this help? Since \(u = \cos(x)\), we can differentiate this equation:

$$\frac{du}{dx} = -\sin(x)$$

The left hand side becomes \(\frac{du}{dx}\) because \(u\) is a function of \(x\) and we had to use the Chain Rule. Now multiply both sides of the equation by \(dx\):

$$du = -\sin(x)dx$$

This can be substituted into the integral to get rid of all the \(x\)s:

$$-\int{\frac{1}{u^2}du}$$

This can be rewritten as

$$-\int{u^{-2}du}$$

by the definition of negative exponents. Integrate this:

$$-u^{-1} + C$$

However, after using u-substitution to integrate, we must undo our substitutions, so let \(u = \cos(x)\):

$$-(\cos(x))^{-1} + C = -\sec(x) + C$$

Example 15: We can find the Maclaurin Series for sine given the Maclaurin Series of cosine. Differentiate the Maclaurin Series for cosine:

$$-\sin(x) = -x + \frac{x^3}{3!} - \frac{x^5}{5!} + ...$$

Now divide both sides of the equation by \(-1\):

$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ...$$

Example 16: Find the value of

$$1 - \frac{4\pi^2}{2!} + \frac{16\pi^4}{4!} - \frac{64\pi^6}{6!} + ...$$

Solution: This looks very similar to the cosine function. Since there are \(\pi\)s everywhere, it looks like it could be \(\cos(\pi)\). Plugging this into the Maclaurin Series gives

$$\cos(\pi) = 1 - \frac{\pi^2}{2!} + \frac{\pi^4}{4!} - \frac{\pi^6}{6!} + ...$$

This is not the same as the series we have to find the value of. However, notice the patterns of the coefficients of the terms. The second term, with a \(2\) in the denominator, has a \(2^2 = 4\) in the numerator, and so on. This suggests that the series we have is equal to \(\cos(2\pi) = 1\). To verify this, substitute \(x \rightarrow 2x\) into the Maclaurin Series for cosine:

$$\cos(2x) = 1 - \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!} + ...$$

Now, substituting \(x = \pi\) gives

$$\cos(2\pi) = 1 - \frac{4\pi^2}{2!} + \frac{16\pi^4}{4!} - \frac{64\pi^6}{6!} + ...$$

Therefore the series is equal to \(1\).